logarithims

[marlo03]

Evaluate the exponential function for three positive values of x, three negative values of x, and at x=0. Show your work. Use the resulting ordered pairs to plot the graph. State the domain and the range of the function.

f(x) = e^(-x )- 1

 

[mmm4444bot]

Evaluate the exponential function for three positive values of x, three negative values of x, and at x=0.

Do you understand the meaning of this instruction ?

Can you pick three positive numbers for x and use them as inputs to function f ?

Where exactly are you stuck, in this exercise ?

PS: Please read the post titled, "Read Before Posting".

 

[marlo03]

I honestly have no clue.I am trying my hardest to understand all of this but I am not getting very far and I have a test on it Monday.

 

[mmm4444bot]

The verb "to evaluate" means to replace all symbols with Real numbers, and then do the arithemtic to arrive at a final number.

You're given an algebraic definition for a function called "f", and you're asked to evaluate it for three positive inputs to this function. You get to pick these three numbers.

f(x) = e^(-x) - 1

I'll pick the following three positive values for x:

x = 1

x = 2

x = 3

When x = 1, we have f(1) = e^(-1) - 1.

Have you learned how to deal with negative exponents? It's a simple property.

e^(-x) = 1/e^x

So e^(-1) = 1/e^1, which is 1/e

So when x = 1, we have the following value for f(x):

f(1) = e^(-1) - 1

f(1) = 1/e - 1

When x = 2, we have:

f(2) = e^(-2) - 1

f(2) = 1/e^2 - 1

When x = 3, we have:

f(3) = e^(-3) - 1

f(3) = 1/e^3 - 1

When x is a negative number, the exponent -x becomes positive.

Let's pick three negative numbers for x:

x = -1

x = -2

x = -3

f(-1) = e^(-[-1]) - 1

f(-1) = e^1 - 1

f(-1) = e - 1

Can you try to figure out f(-2), f(-3), and f(0) ?